Question Number 118384 by mohammad17 last updated on 17/Oct/20
Answered by Dwaipayan Shikari last updated on 18/Oct/20
![∫_0 ^π (x/(1+esinx))dx=∫_0 ^π ((π−x)/(1+esinx))dx=I 2I=π∫_0 ^π (1/(1+esinx))dx 2I=2π∫_0 ^∞ (1/(1+((2te)/(1+t^2 )))).(1/(1+t^2 ))dt I=π∫_0 ^∞ (1/(1+2te+t^2 ))dt I=π∫_0 ^∞ (1/((t+e)^2 −e^2 +1)) I=(π/( (√(1−e^2 ))))[tan^(−1) ((t+e)/( (√(1−e^2 ))))]_0 ^∞ =(π^2 /(2(√(1−e^2 ))))](https://www.tinkutara.com/question/Q118390.png)
$$\int_{\mathrm{0}} ^{\pi} \frac{{x}}{\mathrm{1}+{esinx}}{dx}=\int_{\mathrm{0}} ^{\pi} \frac{\pi−{x}}{\mathrm{1}+{esinx}}{dx}={I} \\ $$$$\mathrm{2}{I}=\pi\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{1}+{esinx}}{dx} \\ $$$$\mathrm{2}{I}=\mathrm{2}\pi\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}{te}}{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${I}=\pi\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{te}+{t}^{\mathrm{2}} }{dt} \\ $$$${I}=\pi\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left({t}+{e}\right)^{\mathrm{2}} −{e}^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}=\frac{\pi}{\:\sqrt{\mathrm{1}−{e}^{\mathrm{2}} }}\left[{tan}^{−\mathrm{1}} \frac{{t}+{e}}{\:\sqrt{\mathrm{1}−{e}^{\mathrm{2}} }}\right]_{\mathrm{0}} ^{\infty} =\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{1}−{e}^{\mathrm{2}} }} \\ $$